GATE Solution 2000 to 2015 GATE SOLUTION to Detailed solution of each question CHEMICAL ENGINEERING GATE SOLUTION

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1 SAMPLE STUDY MATERIAL GATE SOLUTION 000 to 015 Detailed solution of each question CHEMICAL ENGINEERING GATE SOLUTION Subject-wise reducing year CONTENTS

2 GATE Solution 1. Process Calculations Thermodynamics Engineering Fluid Mechanics Mechanical Operations Heat Transfer Mass Transfer Chemical Reaction Engineering Instrumentation and Process Control Plant Design and Economics Chemical Technology Engineering Mathematics Verbal Ability (General Aptitude)

3 1. PROCESS CALCULATIONS (GATE 000 to 015 Question Papers) GATE-015 Q. 1 The schematic diagram of a steady state process is shown below. The fresh feed (F) to the reactor consists of 96 mol% reactant A and 4 mol% inert I. The stoichiometry of the reaction in A C. A part of the reactor effluent is recycled. The molar flow rate of the recycle stream is 0.3F. The product stream P contains 50 mol% C. The percentage conversion of A in the reactor based on A entering the reactor at point 1 in the figure (up to one decimal place) is Q. Adsorption on activated carbon is to be used for reducing phenol concentration in wastewater from 0.04mol/1 to mol/1. The adsorption isotherm at the operating temperature can be expressed as q = 0.05C 1/3 ; where q is the phenol concentration in solid (mol/g solid) and C is the phenol concentration in water (mol/1). The minimum amount of solid (in grams) required per liter of wastewater (up to one decimal place) is GATE-014 Q. 3 A wet solid of 100 kg is dried from a moisture content of 40 wt% to 10 wt%. The critical moisture content is 15 wt% and the equilibrium moisture content is negligible. All moisture basis. The falling rate is considered to be linear. It takes 5 hours to dry the contents are on dry material in the constant rate period. The duration (in hours) of the falling rate period is Q.4 Two elemental gases (A and B) are reacting to form a liquid (C) in a steady state process as per the reaction A + B C. The single-pass conversion of the reaction is only 0% and hence recycle is used. The product is separated completely in pure form. The fresh feed has 49 mol% of A and B each along with mol% impurities. The maximum allowable impurities in the recycle stream is 0 mol%. The amount of purge stream (in moles) per 100 moles of the fresh feed is Q.5 Carbon dioxide (CO) is burnt is presence of 00% excess pure oxygen and the flame temperature achieved is 98K. The inlet streams are at 5ºC. The standard heat of formation (at 5ºC) of CO and CO are 110kJ mol -1 and -390kJ mol -1, respectively. The heat capacities (in J mol -1 K -1 ) of the components are C C po Pco T 3 T

4 Where, T is the temperature in K. The heat loss (in kj) per mole of CO burnt is GATE-013 Common Data for Questions 6 and 7: A reverse osmosis unit treats feed water (F) containing fluoride and its output consists of a permeate stream (P) and a reject stream (R). Let C F, C P, and C R denote the fluoride concentrations in the feed, permeate, and reject streams, respectively. Under steady state conditions, the volumetric flow rate of the reject is 60 % of the volumetric flow rate of the inlet stream, and C F = mg/l and C P = 0.1 mg/l. Q.6 The value of C R in mg/l, up to one digit after the decimal point, is Q.7 A fraction f of the feed is bypassed and mixed with the permeate to obtain treated water having a fluoride concentration of 1 mg/l. Here also the flow rate of the reject stream is 60% of the flow rate entering the reverse osmosis unit (after the bypass). The value of f, up to digits after the decimal point, is GATE-01 Common Data for Questions for 8 and 9 : The reaction A B C D, is carried out in a reactor followed by a separator as shown liq gas liq gas below. Notation: Molar flow rate of fresh B is F FB Molar flow rate of recycle gas is F RG Molar flow rate of A is F A Molar fraction of B in recycle gas is Y RB Molar flow rate of purge gas is F PG Molar flow rate of C is F C Here, F FB = mol/s; F A = 1 mol/s, F B /F A = 5and A is completely converted. Y, the ratio of recycle gas to, purge gas / Q. 8 If 0.3 RB (A) (B) 5 (C) 7 (D) 10 Q. 9 If the ratio of recycle gas to purge gas / (A) 3 8 (B) 5 (C) 1 RG RB F F is RG PG F F is 4 then Y RB is GATE-011 Q. 10 Ammonia is synthesized at 00 bar and 773 K by the reaction 3 (D) 3 4 N 3H NH. The yield of ammonia is 0.45 mol/mol of fresh feed. Flow sheet for the process (along with available compositions) is shown below.

5 The single pass conversion for H in the reactor is 0%. The amount of H lost in the purge as a PERCENTAGE of H in fresh feed is. (A) 10% (B) 0% (C) 45% (D) 55% Q. 11 The following combustion reactions occur when methane is burnt. CH O CO H O 4 CH 3O CO 4H O 4 0% excess air is supplied to the combustor. The conversion of methane is 80% and the molar ratio of CO to CO in the flue gas is 1:3. Assume air to have 80 mol % N and rest O. The O consumed as a PERCENTAGE of O entering the combustor is. (A) 0% (B) 6.5% (C) 80% (D) 83.3% GATE-010 Q. 1 A saturated solution at 30 C contains 5 moles of solute (M.W.=50kg/kmol) per kg of solvent (M.W.=0kg/kmol). The solubility at 100 C is 10 moles of the solute per kg of the solvent. If 10 kg of the original solution is heated to 100 C, then the weight of the additional solute that can be dissolved in it, is (A) 0.5kg (B) 1kg (C) kg (D) 3.34kg Q. 13 The products of combustion of methane in atmospheric air (1% O and 79% N ) have the following composition on a dry basis: Products CO Mole% O.37 CO 0.53 N The ratio of the moles of CH 4 to the moles of O in the feed stream is (A) 1.05 (B) 0.60 (C) 0.51 (D) 0.4

6 GATE-009 Q. 14 Dehydrogenation of ethane, C H g C H g +H g 6 4, is carried out in a continuous stirred tank reactor (CSTR). The feed is pure ethane. If the reactor exit stream contains unconverted ethane along with the products, then the number of degrees of freedom for the CSTR is (1-Mark) (A) 1 (B) (C) 3 (D) 4 Common Data for Question 15 and 16: A flash distillation drum (see figure below) is used to separate a methanol-water mixture. The mole fraction of methanol in the feed is 0.5, and the feed flow rate is 1000 kmol/hr. the feed is preheated in a heater with heat duty Q h and is subsequently flashed in the drum. The flash drum can be assumed to be an equilibrium stage, operating adiabatically. The equilibrium relation between the mole fractions of methanol in the vapor and liquid phases is y=4x. The ration of distillate to feed flow rate is 0.5 Q. 15 The mole fraction of methanol in the distillate is (A) 0. (B) 0.7 (C) 0.8 (D) 0.9 Q. 16 If the enthalpy of the distillate with reference to the feed in 3000 kj/kmol, and the enthalpy of the bottoms with reference to feed is 1000kJ/ kmol, the heat duty of the preheater (Q h in kj/hr) is (A) (B) (C) (D) 10 6 Q. 17 Pure water (stream W) is to be obtained from a feed containing 5wt % salt using a desalination unit as shown below. If the overall recovery of pure water (through stream W) is 0.75 kg/kg feed, then the recycle ratio (R/F) is. (A) 0.5 (B) 0.5 (C) 0.75 (D) 1.0 GATE-008

7 Q. 18 Air (79mole% nitrogen and 1 mole% oxygen) is passed over a catalyst at high temperature. Oxygen completely reacts with nitrogen as shown below. 0.5N 0.5O NO g g g 0.5N O NO g g g The molar ratio of NO to NO in the product stream is :1. The fractional conversion of nitrogen is. (A) 0.13 (B) 0.0 (C) 0.7 (D) 0.40 Q. 19 A 35 wt % Na SO 4 solution in water, initially at 50º C, is fed to a crystallizer at 0º C. The product stream contains hydrate crystals NaSO 4.10H O in equilibrium with a 0 wt % Na SO 4 solution. The molecular weights of Na SO 4 and Na SO 4 10H O are 14 and 3, respectively. The feed rate of 35% solution required to produce 500kg/hr of hydrated crystals is. (A) 403 kg/hr (B) 603 kg/hr (C) 803 kg/hr (D) 1103 kg/hr Q kg/hr of saturated steam at 1 bar (enthalpy kj/kg) is mixed adiabatically with superheated steam at 450ºC and 1 bar (enthalpy kj/kg). The product is superheated steam at 350 C and 1 bar (enthalpy kj/kg). The flow rate of the product is. (A) 711 kg/hr (B) 1111 kg/hr (C)1451 kg/hr (D) 051 kg/hr Q. 1 Carbon black is produced by decomposition of methane: CH C H g g g 4 The single pass conversion of methane is 60 %. If fresh feed is pure methane and 5 % of the methane exiting the reactor is recycled, then the molar ratio of fresh feed stream to recycle stream is (A) 0.9 (B) 9 (C) 10 (D) 90 Common Data for Question, 3, 4: Methane and stream are fed to a reactor in molar ratio 1:. The flowing reactions take place, CH H O CO 4H 4() g () g () g () g CH H O CO 3H 4() g ()() g () g g Where CO is the desired product, CO is the undesired product and H is a byproduct the exit stream has the following composition Species CH 4 H O CO H CO Mole % Q.. The selectivity for desired product relative to undesired product is (A).3 (B) 3.5 (C) 7 (D) 8 Q. 3 The fractional yield of CO is(where fractional yield is defined as the ratio of moles of the desired product formed to the moles that would have been formed if there were no side reactions and the limiting reactant had reacted completely) (A) 0.7 (B) 0.88 (C) 1 (D) 3.5 Q. 4 The fractional conversion of methane is (A) 0.4 (B) 0.5 (C) 0.7 (D) 0.8

8 GATE-007 Linked Answer Questions 5 and 6: A simplified flow sheet is shown in the figure for production of ethanol from ethylene. The conversion of ethylene in the reactor is 30% and the scrubber following the reactor completely separates ethylene (as top stream) and ethanol and water as bottoms. The last (distillation) column gives an ethanol-water azeotrope (90mol% ethanol) as the final product and water as waste. The recycle to purge ratio is 34. The reaction is :C H 4 (g) + H O (g) C H 5 OH(g) Q.5 For an azeotrope product rate of 500 mols/hr the recycle gas flowrate in mols/hr is (A) 30 (B) 40 (C) 100 (D) 1500 Q.6 For the same process, if fresh H O feed to the reactor is 600 mol/hr and wash water for scrubbing is 0% of the condensable coming out of the reactor, the water flowrate in mols/hr from the distillation column as bottoms is (A) 170 (B) 0 (C) 70 (D) 430 Statement for Linked Answer Question 7 and 8: Q.7 44kg of C3H8 is burnt with 1160kg of air (Mol. Wt. = 9) to produce 88kg CO and C H 5O 3CO 4H O 14kg CO 3 8 What is the percent excess air used? (A) 55 (B) 60 (C) 65 (D) 68 Q.8 What is the % carbon burnt? (A) 63.3 (B) 73.3 (C) 83.3 (D) 93.3 Statement for Linked Answer Question 9 and 30: GATE-006 Solvent C is used to extract solute B selectively form 100kh/hr feed mixture A + B in a steady state continuous process shown below. The solubility of C in the raffinate and the solubility of A in the extract are negligible. The extract is distilled to recover B in the bottom product. The overhead product is recycled to the extractor. The loss of solvent in the bottom is compensated by make up solvent S d.. The total flow rate of the solvent stream S going to the extractor is 50kg/hr. the mass fractions (X i s) of some selected streams are indicated in the figure below.

9 Q.9 Distillation bottoms flow rate W and solvent dosing rate S d in kg/hr are (A) W = 50, S d = 50 (B) W = 100, S d = 0 (C) W = 10, S d = 50 (D) W = 50, S d = 10 Q.30 Feed rate E to the distillation column and overhead product rate T in kg/hr are (A) E = 90, T = 40 (B) E = 80, T = 40 (C) E = 90, T 50 (D) E = 45, T = 0 GATE-005 Q.31 A metal recovery unit (MRU) of intake capacity 5000 kg/hr treats a liquid product from a plant and recovers 90% of the metal in the pure form. The unrecovered metal and its associated liquid are sent to a disposal unit along with the untreated product from the plant (See figure below). Find the flow rate (m 6 ) and the weight fraction of the metal (w 6 ). The liquid product flow rate is 7500 kg/hr of composition 0.1 (wt fraction). Assume steady state. m 7500 kg / hr, w 0.0 (B) m kg / hr, w (A) 6 6 m 4500 kg / hr, w (D) m kg / hr, w (C) 6 6 :::::::::::::::::::::Sample File Continue:::::::::::::::::::::::::

10 PROCESS CALCULATIONS: Answer key & Detailed explanations % 6.4g to 3.0 to mg/L 0.86 B A A B C D C C D B B C D B C A C C D D C D A A C C D D C D A B C SAMPLE SAMPLE SAMPLE Detailed Solutions & Explanations Explanations-015 ANS -1 : 45.5% EXP: Basis: 100 kmol of fresh feed (F). It contains 96 kmol A and 4 kmol inert. A in outlet of reactor = A outlet from separator =Sum of A in product and recycle

11 So, A in to the reactor at point (1) - A in outlet of reactor % conversion of A = 100 A in to the reactor at point (1) % ANS- : 6.4g EXP: Phenol removed from warts water = = 0.03 mol Let q be the mol of phenol in 1 g solid in saturation. q mol phenol = 1 g solid 0.03 mol phenol 0.03 g solid required (i) q q 0.05C q 1/3 1/3 0.05(0.008) [Given] q By equation (1), solid required g q Explanations Wet solid = 100kg Initial Moisture content = 40wt% (x i ) Final moisture content = 10wt% (x c ) Equivalent moisture content = 0 (x * ) or (x e ) All the moisture contents are on dry basis. (Given) Let Ws be the weight of bone dry solid. Moisture content is defined as kg moisture kg dry solid To make all the moisture contents satisfy the above form, they must be divided form Ws. Constant drying time = 5hr (Given) For the constant drying rate Regime Ws t () x x AN c i c c Where N C Constant drying rate from the above formula, we can calculate AN C by putting. u.15 xi, xc ws ws t c = 5hr AN C = 1/0 (calculated value)

12 For the falling rate regime, drying rate is linear (Given) We can use the formula for falling rate period Ws * xc x t f () xc x Ln AN x x C By putting the values t 1.(0)(.15)(1.5) hr t Ln f f f * * to to Feed (F) C = mg/l F RO Explanations-013 Reject (R) Permeate stream (P) C P = 0.1 mg/l Mass balance F = P + R Given: R = 0.6 F (i) F = P F P = 0.4F (ii) Mass balance on fluoride content : FCF PCP RCR From equation (i) and (ii) F = 0.4 F F CR 7. C R mg/ L 0.6 Given: R = 60% of volumetric flow rate of the inlet stream Therefore, P = 0.4 F (1 f ) (i) Fluoride content balance on by pass stream; f F [P f F] C P f F = P + f F CP 1 (Given) From equation (i) f F = 0.4 F(1 f ) + f F f = 0.4 (1 f ) + f f 0.4 /

13 8. (B) GATE Solution 000 to 015 Explanations - 01 FB Given: FFB mol / sec 5 mol / sec F FA 1 mol / sec A is completely converted Assume separator separatesall C F 1 mol / sec Overall Material Balance across the dotted circle A FA FFB FC FPG FPG FA FFB - FC 1-1 mol / sec Material Balance for component B at the point (1) C F F Y F F B FB RB RG RG FB - F Y RB FB F F Y 0.3 given so F 10 mol / sec RB FRG F PG RG 9. (A) 4given PG RG F PG from Question 50 F RG 4 F 4 8 mol / sec. PG Material Balance at point (1) FB FFB YRB FRG 10.(A) Y RB FB - FFB 5 3 F 8 8 RG FB 5 mol / sec given FFB mol / sec Explanations - 011

14 N 3H NH 3 Basis: 100 moles of feed As yield of ammonia =0.45 mol/mol of fresh feed Molesof NH 3 Produced = =45mol 135 moles of H 3 required 45 mole Single pass conversion (0%) moles of H entering mole Moles of H Coming out from separator = Moles of H in fresh feed mol 135 In recycle moles of H mol 0.4 In purge moles of H = = 7.5 mol Amount of H lost in the purge as a % of H in fresh feed CH O CO H O...(1) ; 11. (B) % 75 CH 3O CO 4 H O...(); 4 = Extent of reaction. a Let assume 1 b By Stoichiometry O required = mol 0% excess ; O Supplied 1..4mol Air; 80% N and 0% O 0.8 So N supplied mole 0. 80% Conversion so, CH 4 at the exit of combustor =1(1-0.8) = 0. Methane mole balance;

15 mole of CO mole of CO From above two equation (3) and (4) GATE Solution 000 to (4) O consumed 3 O sup plied = Explanations (C) At 30ºC solution contains 5 moles of solute C sol. 5mol solute kg solvent kg solute 5mol solute 1k mol solute 50 kg solute kg solute Means Csol. 0.5 kg solvent kg solvent 1000 mol solute k mol solute kg solvent 0.5 kg solute kg solute Csol kg solvent kg solution 10 Kg solution contains =10 0. = kg solute and 8 kg solvent 10 mol solute At100º C Csol kg solvent So the additional solute =4 = kg solute CH O CO H O 13. (D) 4 10 mol solute 1k mol solute 50 kg solute = 4 kg solute kg solvent 1000 mol solute k mol solute 3 CH 4 O CO H O From above two equation Mole of CH 4 burnt = mole of CO + mole of CO = = moles Moles of O entered 1moles O moles of N mol O 79 mol N Ratio ::::::::::::::::::::::::::::::::::::::Sample File:::::::::::::::::::::::::::::::::::::::: Highest Result in GATE 015 Rank 1,, 7, 8.Total 39 Ranks under AIR 100 To Buy Postal Correspondence Package call at